Visible light travels in air, which has an index of refraction of 1.0. It is incident on plastic which is covered by a coating (such that li

Question

Visible light travels in air, which has an index of refraction of 1.0. It is incident on plastic which is covered by a coating (such that light hits the coating before the plastic). The coating has an index of refraction of 1.6 and is 0.50 microns thick. The ray reflected off the air-coating interface and the ray reflected off the coating-plastic interface experience interference. What frequencies could the light be

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Acacia 2 weeks 2021-09-05T12:49:42+00:00 1 Answers 0 views 0

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    2021-09-05T12:51:30+00:00

    Answer:

      λ = 0.64 10⁻⁶ m = 640 nm

    Explanation:

    The two reflected rays experience constructive interference, if we can see them, so we can one condition for interference, but let’s see two phenomena that occur

    * A ray when it is reflected by a surface of major spare part index has a phase change of 180º, these passes at the air-plastic interface

    * A ray when passing a material with a refractive index changes its wavelength

           λ= λ₀ / n

    taking into account these facts the condition in constructive interference is

                 2 n t = (m + ½) λ

                 λ= 2 n t / (m + ½)

    if we suppose that we have the first inference m = 0

                λ = 2 1.6 0.5 10⁻⁶ (0 + ½)

                λ = 3.2 10⁻⁶ m

    this wavelength is in the infrared

    suppose an interference of m = 1

              λ = 1.6 10⁻⁶ / (1 + 1/2)

              λ = 1.06 10-6 m

    m = 2

             λ = 1.6 10⁻⁶ / (2 + 1.5)

             λ = 0.64 10⁻⁶ m

    this wavelength corresponds to the visible range (640 nm) orange

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